Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int N = 200000 + 3;
vector<int> l(N), r(N);
int n;
double sum[N];
double intersec(double x1, double y1, double x2, double y2) {
  return x1 - y1 * (x1 - x2) / (y1 - y2);
}
int binary_search(vector<int>& arr, double v) {
  int ll = 0, rr = n - 1;
  while (ll <= rr) {
    int m = ll + (rr - ll) / 2;
    double t = (double)arr[m];
    if (t > v)
      rr = m - 1;
    else if (abs(t - v) < eps)
      return m;
    else
      ll = m + 1;
  }
  return ll;
}
int main() {
  int sy, a, b;
  scanf("%d%d%d", &(sy), &(a), &(b));
  scanf("%d", &(n));
  for (int i = 0; i < (n); ++i) {
    scanf("%d%d", &(l[i]), &(r[i]));
    sum[i] = double(r[i] - l[i]);
    if (i > 0) sum[i] += sum[i - 1];
  }
  int q;
  scanf("%d", &(q));
  for (int i = 0; i < (q); ++i) {
    int x, y;
    scanf("%d%d", &(x), &(y));
    double L = intersec((double)x, (double)y, (double)a, (double)sy);
    double R = intersec((double)x, (double)y, (double)b, (double)sy);
    int ll = binary_search(l, L);
    int rr = binary_search(r, R);
    double len = 0.0;
    if (ll > rr)
      len = R - L;
    else {
      if (ll == n)
        len = max((double)r[rr - 1] - L, 0.0);
      else {
        if (ll > 0) len += max((double)r[ll - 1] - L, 0.0);
        if (rr < n) len += max(R - (double)l[rr], 0.0);
        rr--;
        if (ll <= rr) len += sum[rr] - sum[ll] + (double)(r[ll] - l[ll]);
      }
    }
    len = len * (double)(b - a) / (R - L);
    printf("%lf\n", len);
  }
  return 0;
}